# Thread: Ordinary differential equations

1. ## Ordinary differential equations

Hi there. Well, I have some doubts about this exercise, I think I've solved it, but I wanted your opinion, which always help. So, it says:

There is no general method for solving the homogeneous general equation of second order
$y''+P(x)y'+Q(x)y=0$ (1)

But if we already know a solution $y_1(x)\neq 0$, then we always can find a second solution linearly independent

$y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx$
Demonstrate that this functions form a basis for the space of solutions of the differential equation (1).

So, what I did is simple. I know that if the solution is linearly independent, then the Wronskian $w(y_1,y_2,x)$ must be zero. So, I just made the calculus for the wronskian:

$\left| \begin{matrix}{y_1}&{y_2}\\{y_1'}&{y_2'}\end{matri x} \right|=\left| \begin{matrix}{y_1}&{y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx}\\{y_1'}&{y_1'(x) \int \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2}dx+y_1(x) \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2(x)}dx}\end{matrix} \right|=\displaystyle\frac{e^{- \int P(x)dx}}{y_1(x)}\neq0$

Is this right?

Bye there.

2. Your approach is fine; however I don't think your determinant computation is correct. You've set it up properly, but I get just

$e^{-\int P(x)\,dx}$

for the determinant.

3. You're right! thanks.

4. You're welcome!

5. This is pretty much like the same. It gives me a differential equation, and one solution and asks me for the entire solution. Did I do it well?

$(1-2x)y''+2y'+(2x-3)y=0,y_1=e^x$
$y_2=e^x \int \displaystyle\frac{e^{-\int 2dx}}{e^{2x}}dx=\displaystyle\frac{e^{-3x}}{-4}$

Then $y(x)=Ae^x-B\displaystyle\frac{e^{-3x}}{4}$

Is that okey? I'm not sure, because I've tried to corroborate the solution with wolfram alpha, but it gives something different.

Here:
&#40;1-2x&#41;y&#39;&#39;&#43;2y&#39;&#43;&#40;2x-3&#41;y&#61;0 - Wolfram|Alpha

6. You haven't normalized the DE to look like (1) in the OP. You have to do that before applying the variation of parameters formula.

7. Sorry, I've missed that part, how should I do that?

8. Just divide the entire DE by 1-2x to get the coefficient of y'' equal to 1.